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\title{
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\textsc{中国科学院大学   计算机与控制学院} \\ [25pt] % Your university, school and/or department name(s)
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\huge 随机过程第八次作业 \\ % The assignment title
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}
\author{黎吉国&201618013229046} % Your name
%\date{\normalsize Nov 8,2016}

\begin{document}

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\newpage
\section{泊松分布}
\begin{enumerate}
  \item 定义
  \begin{itemize}
    \item $N(0)=0$
    \item 独立增量
    \item 平稳增量
    \item 无穷小的时间间隔中基本只会发生一次
  \end{itemize}
  \item 基本性质
  \begin{itemize}
    \item $E(N(t))=\lamda t$
    \item $Var(N(t))=\lamda t$
    \item $\phi_N (t,v)=\exp(\lambda t(e^{jv}-1))$
  \end{itemize}
  \item
设在$[0,t]$内事件$A$已经发生了$n$次，且$0<s<t$,对于$0<k<n$,求$P\{ X(s)=k|X(t)=n \}$\\
\textbf{解：}\\
\[
\begin{split}
  P\{ X(s)=k|X(t)=n \}&=\frac{P\{ X(s)=k,X(t)=n \}}{P\{ X(t)=n \}}\qquad \text{条件概率}\\
  &=\frac{P\{ X(s)=k,X(t)-X(s)=n-k \}}{P\{ X(t)=n \}}\\
  &=C_n^k(\frac{s}{t})^k(1-\frac{s}{t})^{n-k}
\end{split}
\]
\item
设$X_1,X_2,...,X_n$是统计独立且拥有相同的分布的一组随机变量，令随机变量$Y=\Sigma_{k=1}^{n}X_k$，证明，若$X_k$服从
参数为$\lambda_k$的泊松分布吗，则$Y$必须服从参数为$\lambda_Y$的泊松分布，其中$\lambda_Y=\Sigma_{k=1}^{n}\lambda_k$\\
\textbf{证明}：（随机过程的复合，适合用特征函数来分析）\\
所$X_k$服从参数为$\lambda_k$的泊松分布，则特征函数为（见随机过程（刘次华）P31）
\[ \phi_{X_k}(v,t)=\exp{\{  \lambda_k t(e^{jv}-1)\}}\]
又由于$X_1,X_2,\ldots,X_n$相互独立，因此随机变量$Y$的特征函数为
\[
\begin{split}
  \phi_Y(v) &= \prod_{k=1}^n \phi_{X_k}(v)\\
  &= \exp{\{ \Sigma_{k=1}^{n} \lambda_k t(e^{jv}-1) \}}\\
  &= \exp{\{ \lambda_Y (e^{jv}-1) \}}
\end{split}
\]
得证。
\item 复合泊松分布\\
已知复合泊松分布为
\[ Y_t=\Sigma_{k=1}^{N(t)}X_n,t\ge 0 \]
其中$X_n$服从以下分布
\[
P(X<x)=
\begin{cases}
  1-e^{-ux} &\mbox x\ge 0\\
  0 &\mbox x<0
\end{cases}
\]
求复合泊松分布的均值，方差，和特征函数。\\
\text{解：}\\
已知复合泊松过程的特征函数为(见随机过程（刘次华）P39)
\[ \phi_Y(v,t)=\exp{\{ \lambda t( \phi_X(v,t)-1 ) \}} \]
\[ E(Y_t)=\lambda t E(X) \]
\[ Var(Y_t)=\lambda t E(X^2) \]
由$X_n$的分布可得
\[
\begin{split}
  E(X)&=\frac{1}{u}\\
  E(X^2)&=\frac{2}{u^2}\\
  \phi _X(v)&= \frac{u}{u-jv}
\end{split}
\]
可得
\[
\begin{split}
E(Y_t)&=\frac{\lambda t}{u}\\
Var(Y_t)&=\frac{2\lambda t}{u^2}\\
\phi_Y(v,t)&=\exp{\{ \frac{j\lambda v t}{u-jv} \}}
\end{split}
\]

\end{enumerate}

\end{document}
